In Vignéras' book on quaternion algebras, she uses the properties of the zeta functions to deduce some results on the classification of quaternion algebras. The zeta function attached to a quaternion algebra $H$ over a global field $F$ (say $\mathbb Q$ to lighten notations) is related to the Dedekind zeta function by $$\zeta_H(s/2) = \zeta_F(s) \zeta_F(1-s) \prod_{p\in \mathrm{Ram}(H)} (1-p^{1-s})$$ We know $Z_H(s)$ has a simple pole at $0$ and $1$ if $H$ is a field (in particular for $F$), and is holomorphic elsewhere. But the above finite product has a zero of order $\mathrm{Ram}(H)$ at $1$. Taking $s=1$ above, we get two poles from $\zeta_K$ and this zero, hence a zero of order $\mathrm{Ram}(H)-2$.
How then do we deduce the equivalence that $H=M_2(F)$ if and only if $H_v=M_2(F_v)$ at each place $v$ of $F$?
I understand that if $Ram(H)=\varnothing$, then there is a new pole at $1/2$ and we cannot have a field. But otherwise, we get a zero and there is no obstruction for $H$ to be a field (but no obligation either!)